You are given an array A of length N.

You have to partition the elements of the array into some [subsequences] such that:
Each element A_{i} (1 ≤ i ≤ N) belongs to exactly one subsequence.
The [mean] of the mean of subsequences is maximised.

Formally, let $S_{1}, S_{2}, \dots, S_{K} denote K$ subsequences of array A such that each element A_{i} (1 ≤ i ≤ N) belongs to exactly one subsequence S_{j} (1 ≤ j ≤ K).
Let X_{j} (1 ≤ j ≤ K) denote the mean of the elements of subsequence S_{j}. You need to maximise the value $\frac{\sum_{j=1}^K X_{j}}{K}$.

Print the maximum value. The answer is considered correct if the relative error is less than 10^{-6}.

------ Input Format ------

• First line will contain T, number of test cases. Then the test cases follow.
• First line of each test case consists of a single integer N - denoting the length of array A.
• Second line of each test case consists of $N space-separated integers A_{1}, A_{2}, \ldots, A_{N} -$ denoting the array A.

------ Output Format ------

For each test case, output in a single line, the maximum possible mean of mean of subsequences. The answer is considered correct if the relative error is less than 10^{-6}.

------ Constraints ------

$1 ≤ T ≤ 1000$
$2 ≤ N ≤ 10^{5}$
$1 ≤ A_{i} ≤ 10^{6}$
- Sum of $N$ over all test cases does not exceed $3 \cdot 10^{5}$.

----- Sample Input 1 ------
3
2
10 20
3
1 2 3
5
50 50 50 50 50

----- Sample Output 1 ------
15
2.25
50

----- explanation 1 ------
Test Case $1$: We can partition the array in $2$ ways - $(\{10\}, \{20\})$ or $(\{10,20\})$. In both cases, mean of mean of subsequences comes out to be $15$.

Test Case $2$: The optimal partition is $(\{1, 2\}, \{3\})$.
- Mean of first subsequence $= \frac{1+2}{2} = 1.5$.
- Mean of second subsequence $= \frac{3}{1} = 3$.

Thus, mean of mean of subsequences is $\frac{1.5+3}{2} = \frac{4.5}{2} = 2.25$.

Test Case $3$: Any partition will yield the mean of mean of subsequences as $50$.

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